Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient/Proof 2

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Theorem

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Then:

\(\ds \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k}\) \(=\) \(\ds \binom {2 n + 1} 2 2^2 B_2 + \binom {2 n + 1} 4 2^4 B_4 + \binom {2 n + 1} 6 2^6 B_6 + \cdots\)
\(\ds \) \(=\) \(\ds 2 n\)

where $B_n$ denotes the $n$th Bernoulli number.


Proof

The proof proceeds by induction.

For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:

$\ds \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k} = 2 n$


Basis for the Induction

$\map P 1$ is the case:

\(\ds \binom {2 \times 1 + 1} 2 2^2 B_2\) \(=\) \(\ds \frac {2 \times 3} 2 \times 2^2 \times \frac 1 6\) Binomial Coefficient with Two, Definition of Bernoulli Numbers
\(\ds \) \(=\) \(\ds 2\) after simplification
\(\ds \) \(=\) \(\ds 2 \times 1\)


Thus $\map P 1$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P r$ is true, where $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true.


So this is the induction hypothesis:

$\ds \sum_{k \mathop = 1}^r \dbinom {2 r + 1} {2 k} 2^{2 k} B_{2 k} = 2 r$


from which it is to be shown that:

$\ds \sum_{k \mathop = 1}^{r + 1} \dbinom {2 \paren {r + 1} + 1} {2 k} 2^{2 k} B_{2 k} = 2 \paren {r + 1}$


Induction Step

This is the induction step:


\(\ds \sum_{k \mathop = 1}^{r + 1} \dbinom {2 \paren {r + 1} + 1} {2 k} 2^{2 k} B_{2 k}\) \(=\) \(\ds \sum_{k \mathop = 1}^r \dbinom {2 \paren {r + 1} + 1} {2 k} 2^{2 k} B_{2 k} + \dbinom {2 \paren {r + 1} + 1} {2 \paren {r + 1} } 2^{2 \paren {r + 1} } B_{2 \paren {r + 1} }\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^r \dbinom {2 r + 3} {2 k} 2^{2 k} B_{2 k} + \paren {2 r + 3} 2^{2 r + 2} B_{2 r + 2}\) Binomial Coefficient with Self minus One and simplification



So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\ds \forall n \in \Z_{>0}: \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k} = 2 n$